Calculus of variations — functionals, functional derivative: Worked examples

This child page of the parent topic does one thing: it drills the Euler–Lagrange machine (from the parent) against every kind of case it can meet. We start by listing the cases, then work an example for each cell.

Before line one, three reminders in plain words (nothing here is assumed — it is all in the parent):

The scenario matrix

Every problem in this topic falls into one of these shapes. The last column names the example that covers it.

#	Case class	What is special about it	Covered by
C1	$L$ has no $y$ ( $\partial L / \partial y = 0$ )	E–L collapses to " $L_{y^{'}} =$ const"	Ex 1
C2	$L$ has no $x$ explicitly	Use Beltrami (avoid $y^{''}$ )	Ex 2
C3	$L$ has no $y^{'}$ ( $\partial L / \partial y^{'} = 0$ )	E–L becomes algebraic, no ODE	Ex 3
C4	Full $L (x, y, y^{'})$ — all slots present	Total-derivative chain rule matters	Ex 4
C5	Degenerate / limiting input ( $ω \to 0$ , $c \to \infty$ , straight-line limit)	Check the machine gives the sane limit	Ex 5
C6	Free endpoint — boundary term survives	Natural boundary condition appears	Ex 6
C7	Constraint (fixed area/length)	Lagrange multiplier enters $L$	Ex 7
C8	Word problem (real-world: fastest slide)	Model → identify $L$ → solve	Ex 8
C9	Exam twist (sign trap / higher derivative)	Where students lose marks	Ex 9

Read each example's tag [Cn] to see which cell it fills.

Ex 1 — No $y$ in $L$ `[C1]`

Worked example Shortest path, done carefully

Minimise $J [y] = \int_{0}^{1} 1 + y^{'2} d x$ with $y (0) = 0, y (1) = 2$ . Here $L = 1 + y^{'2}$ .

Forecast: guess the answer before reading. Shortest wire between two points → what shape? (Say it out loud.)

Check which cell. $L$ contains $y^{'}$ but no bare $y$ . Why this step? If $\partial L / \partial y = 0$ , E–L loses its first term and becomes easy.
$\frac{\partial L}{\partial y ^{'}} = \frac{y ^{'}}{1 + y ^{'2}}$ . Why? Differentiate $1 + y^{'2}$ treating $y^{'}$ as the variable (chain rule: $\frac{1}{2} (1 + y^{'2})^{- 1/2} \cdot 2 y^{'}$ ).
E–L: $0 - \frac{d}{d x} \frac{y ^{'}}{1 + y ^{'2}} = 0 \Rightarrow \frac{y ^{'}}{1 + y ^{'2}} = c$ . Why? A derivative equal to zero means the inside is constant.
Solve for $y^{'}$ : square → $y^{'2} = c^{2} (1 + y^{'2}) \Rightarrow y^{'2} (1 - c^{2}) = c^{2} \Rightarrow y^{'} = \frac{c}{1 - c ^{2}} \equiv m$ (constant). Why? Constant slope.
Integrate: $y = m x + b$ . Fit endpoints: $b = 0$ , $m = 2$ . So $y = 2 x$ .

Verify: the straight segment from $(0, 0)$ to $(1, 2)$ has length $1^{2} + 2^{2} = 5 \approx 2.236$ . Any bend adds length. ✓ Matches the forecast — a straight line.

Figure — Calculus of variations — functionals, functional derivative

Ex 2 — No explicit $x$ → Beltrami `[C2]`

Worked example Minimal surface of revolution (catenary)

Minimise $J [y] = \int_{a}^{b} y 1 + y^{'2} d x$ , $y > 0$ . So $L = y 1 + y^{'2}$ — a soap-film area.

Forecast: hanging-chain shape or parabola? Guess.

Spot the cell: no explicit $x$ in $L$ . Why this step? Then Beltrami $L - y^{'} L_{y^{'}} = const$ dodges the ugly $y^{''}$ .
$L_{y^{'}} = \frac{y y ^{'}}{1 + y ^{'2}}$ . Why? Only $1 + y^{'2}$ depends on $y^{'}$ ; product with the constant-in- $y^{'}$ factor $y$ .
Beltrami: $y 1 + y^{'2} - y^{'} \cdot \frac{y y ^{'}}{1 + y ^{'2}} = \frac{y}{1 + y ^{'2}} = c$ . Why? Combine over common denominator; the numerator collapses to $y$ .
Rearrange: $y^{2} = c^{2} (1 + y^{'2}) \Rightarrow y^{'} = \frac{y ^{2}}{c ^{2}} - 1$ . Why? Isolate $y^{'}$ to get a separable ODE (see Ordinary differential equations).
Separate & integrate: $\int \frac{d y}{y ^{2} / c ^{2} - 1} = \int d x$ gives $y = c cosh (\frac{x - x _{0}}{c})$ . Why? $\frac{d}{d u} cosh u = sinh u$ and $cosh^{2} - sinh^{2} = 1$ make it fit.

Verify: put $y = c cosh \frac{x}{c}$ (take $x_{0} = 0$ ). Then $y^{'} = sinh \frac{x}{c}$ and $\frac{y}{1 + y ^{'2}} = \frac{c cosh}{1 + sinh ^{2}} = \frac{c cosh}{cosh} = c$ . Constant ✓ — Beltrami satisfied. This is the geometry behind soap films.

Ex 3 — No $y^{'}$ in $L$ : E–L is algebraic `[C3]`

Ex 4 — Full $L (x, y, y^{'})$ : the total-derivative trap `[C4]`

Worked example All three slots present

$J [y] = \int_{0}^{1} (y^{'2} + 12 x y) d x$ , $y (0) = 0, y (1) = 1$ . $L = y^{'2} + 12 x y$ has $x$ , $y$ and $y^{'}$ .

Forecast: degree of the resulting polynomial $y (x)$ ? Guess.

Cell: every slot used — the total $x$ -derivative must include $x$ moving. Why this step? This is where beginners wrongly use a partial derivative.
$\frac{\partial L}{\partial y} = 12 x$ , $\frac{\partial L}{\partial y ^{'}} = 2 y^{'}$ . Why? Partial derivatives, one slot at a time.
$\frac{d}{d x} (2 y^{'}) = 2 y^{''}$ . Why? Here $L_{y^{'}} = 2 y^{'}$ has no explicit $x$ or $y$ , so the total derivative is just $2 y^{''}$ (the general chain rule $L_{y^{'} x} + L_{y^{'} y} y^{'} + L_{y^{'} y^{'}} y^{''}$ reduces to the last term).
E–L: $12 x - 2 y^{''} = 0 \Rightarrow y^{''} = 6 x$ . Why? Set the E–L expression to zero.
Integrate twice: $y^{'} = 3 x^{2} + A$ , $y = x^{3} + A x + B$ . Endpoints: $B = 0$ ; $1 = 1 + A \Rightarrow A = 0$ . So $y = x^{3}$ .

Verify: $y = x^{3} \Rightarrow y^{''} = 6 x$ ✓ and $y (0) = 0, y (1) = 1$ ✓. Cubic, as the forecast should have been.

Ex 5 — Degenerate / limiting input `[C5]`

Worked example The oscillator energy as

$ω \to 0$ From the parent: $J [y] = \int_{0}^{1} (\frac{1}{2} y^{'2} + \frac{1}{2} ω^{2} y^{2}) d x$ gives E–L $y^{''} = ω^{2} y$ . Now push the limit $ω \to 0$ .

Forecast: what shape should a taut string with no stiffness prefer? Guess.

Cell: limiting/degenerate parameter. Why this step? A good machine must give sane answers at extreme inputs.
General solution of $y^{''} = ω^{2} y$ (with $ω > 0$ ): $y = P cosh (ω x) + Q sinh (ω x)$ . Why? Positive coefficient → hyperbolic (growing/decaying) modes.
Set $ω \to 0$ : $cosh (ω x) \to 1$ , $\frac{sinh ( ω x )}{ω} \to x$ . So $y \to$ (constant) $+$ (constant) $\cdot x$ = straight line. Why? The equation degenerates to $y^{''} = 0$ , whose solutions are lines.
Sanity with Ex 1: no restoring term ( $ω = 0$ ) leaves just $\frac{1}{2} y^{'2}$ — a "shortest-slope" energy → straight line. ✓

Verify (numeric): take endpoints $y (0) = 1, y (1) = 3$ . Limit answer $y = 1 + 2 x \Rightarrow y (0.5) = 2$ . Now the finite case $ω = 0.001$ with the same endpoints gives $y (0.5) \approx 2.0000$ to 4 dp — matches the limit. ✓

Ex 6 — Free endpoint → natural boundary condition `[C6]`

Worked example One end is loose

Minimise $J [y] = \int_{0}^{1} (\frac{1}{2} y^{'2} - y) d x$ with $y (0) = 0$ but $y (1)$ free (not prescribed).

Forecast: with the right end free, what slope will nature choose there? Guess.

Cell: free endpoint. Why this step? The IBP boundary term $[L_{y^{'}} η]_{0}^{1}$ no longer dies at $x = 1$ , because $η (1)$ is not forced to zero.
Interior E–L: $L_{y} = - 1$ , $L_{y^{'}} = y^{'}$ , so $- 1 - y^{''} = 0 \Rightarrow y^{''} = - 1$ . Why? Same E–L inside the interval.
Boundary term at the free end must vanish for all $η (1)$ : $L_{y^{'}}_{x = 1} = 0 \Rightarrow y^{'} (1) = 0$ . Why? This is the natural boundary condition — the machine hands you the missing endpoint data.
Solve $y^{''} = - 1$ : $y = - \frac{1}{2} x^{2} + A x + B$ . Use $y (0) = 0 \Rightarrow B = 0$ ; $y^{'} (1) = - 1 + A = 0 \Rightarrow A = 1$ . So $y = - \frac{1}{2} x^{2} + x$ .

Verify: $y^{'} = - x + 1$ , so $y^{'} (1) = 0$ ✓ (flat slope at the loose end, as forecast) and $y (0) = 0$ ✓. See Functional analysis for why free ends generate conditions instead of requiring them.

Ex 7 — Constraint via Lagrange multiplier `[C7]`

Worked example Longest area under fixed length (isoperimetric)

Among curves $y (x)$ on $[0, 1]$ with $y (0) = y (1) = 0$ and fixed length $\int_{0}^{1} 1 + y^{'2} d x = ℓ$ , maximise the enclosed area $\int_{0}^{1} y d x$ .

Forecast: circle-arc, parabola, or catenary? Guess.

Cell: constrained problem. Why this step? Attach the constraint with a multiplier $λ$ (see Constrained optimization & Lagrange multipliers): form $L = y + λ 1 + y^{'2}$ .
$L_{y} = 1$ , $L_{y^{'}} = \frac{λ y ^{'}}{1 + y ^{'2}}$ . Why? Ordinary partials of the augmented integrand.
E–L: $1 - \frac{d}{d x} \frac{λ y ^{'}}{1 + y ^{'2}} = 0 \Rightarrow \frac{λ y ^{'}}{1 + y ^{'2}} = x - x_{0}$ . Why? Integrate the E–L once (the $1$ integrates to $x$ ).
Solve for $y^{'}$ and integrate → $(x - x_{0})^{2} + (y - y_{0})^{2} = λ^{2}$ : a circular arc. Why? The relation says the tangent direction rotates uniformly with $x$ -displacement — the signature of a circle.

Verify: on a circle of radius $R = λ$ , $\frac{y ^{'}}{1 + y ^{'2}} = sin θ$ where $θ$ is the tangent angle; and $R sin θ$ = horizontal offset $x - x_{0}$ ✓. The classical isoperimetric answer is the circle. Forecast: circle-arc.

Ex 8 — Word problem: fastest slide `[C8]`

Worked example Brachistochrone, framed from words

A bead slides frictionlessly from $(0, 0)$ to $(x_{1}, y_{1})$ under gravity $g$ , starting at rest. Find the wire shape giving the shortest travel time. (Full context: Brachistochrone problem.)

Forecast: straight ramp, or a curve that dives steeply first? Guess.

Model. Speed from energy: $\frac{1}{2} v^{2} = g y \Rightarrow v = 2 g y$ (measure $y$ downward). Time $= \int \frac{d s}{v}$ with $d s = 1 + y^{'2} d x$ . Why this step? Turn "time" into a functional so E–L applies.
So $J [y] = \int_{0}^{x_{1}} \frac{1 + y ^{'2}}{2 g y} d x$ , giving $L = \frac{1 + y ^{'2}}{2 g y}$ . No explicit $x$ → Beltrami. Why? Same trick as Ex 2.
Beltrami: $L - y^{'} L_{y^{'}} = \frac{1}{2 g y 1 + y ^{'2}} = const$ . Why? Algebra identical in form to Ex 2's collapse.
Rearrange: $y (1 + y^{'2}) = const = k$ . Why? Square and clear denominators.
This ODE is solved by a cycloid: $x = \frac{k}{2} (ϕ - sin ϕ), y = \frac{k}{2} (1 - cos ϕ)$ . Why? Substituting $y = \frac{k}{2} (1 - cos ϕ)$ satisfies $y (1 + y^{'2}) = k$ .

Verify: at $ϕ = π$ , $y = \frac{k}{2} (1 - (- 1)) = k$ ; then $y^{'} = 0$ (bottom of the arc) so $y (1 + y^{'2}) = k \cdot 1 = k$ ✓. The bead dives first — forecast confirmed. This links to Lagrangian mechanics where the same $L$ -machinery runs mechanics.

Ex 9 — Exam twist: sign trap & a $y^{''}$ in $L$ `[C9]`

Worked example Second-derivative functional (Euler–Poisson)

Minimise $J [y] = \int_{0}^{1} (\frac{1}{2} y^{''2}) d x$ — the "bending energy" of a beam — with $y (0) = 0, y^{'} (0) = 0, y (1) = 1, y^{'} (1) = 0$ . Note $L$ depends on $y^{''}$ !

Forecast: plain E–L is not enough here. What order will the governing equation be? Guess.

Cell / twist: $L$ has $y^{''}$ , so the ordinary E–L is incomplete. Why this step? Repeating the IBP twice adds a $+ \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}}$ term (Euler–Poisson equation): $\frac{\partial L}{\partial y} - \frac{d}{d x} \frac{\partial L}{\partial y ^{'}} + \frac{d ^{2}}{d x ^{2}} \frac{\partial L}{\partial y ^{''}} = 0$ .
Here $L_{y} = 0, L_{y^{'}} = 0, L_{y^{''}} = y^{''}$ . So the equation is $\frac{d ^{2}}{d x ^{2}} (y^{''}) = 0 \Rightarrow y^{(4)} = 0$ . Why? Only the $y^{''}$ slot is nonzero; differentiate it twice.
Sign trap: the sign is plus on the $y^{''}$ term (two IBPs = two minus signs = a plus). Dropping a sign here is the classic exam error. Why? Each IBP flips the sign; even number of flips → positive.
Solve $y^{(4)} = 0$ : cubic $y = a x^{3} + b x^{2} + c x + d$ . Apply the four conditions: $y (0) = 0 \Rightarrow d = 0$ ; $y^{'} (0) = 0 \Rightarrow c = 0$ ; $y (1) = a + b = 1$ ; $y^{'} (1) = 3 a + 2 b = 0$ . Solve: $b = 3, a = - 2$ . So $y = - 2 x^{3} + 3 x^{2}$ .

Verify: $y (0) = 0, y^{'} (0) = 0$ ✓. $y (1) = - 2 + 3 = 1$ ✓. $y^{'} (x) = - 6 x^{2} + 6 x \Rightarrow y^{'} (1) = 0$ ✓. And $y^{(4)} = 0$ ✓ (it's a cubic). Fourth-order equation, as forecast. This is the smooth-blend curve engineers call a Hermite cubic.

Recap of the matrix

Recall Did we hit every cell?

Which example proves "no $y$ ⇒ $L_{y^{'}} =$ const"? ::: Ex 1 (straight line) Which uses Beltrami because $L$ lacks explicit $x$ ? ::: Ex 2 (catenary) and Ex 8 (cycloid) Which turns E–L into pure algebra? ::: Ex 3 ( $y = x$ ) Which needs the full total-derivative chain? ::: Ex 4 ( $y = x^{3}$ ) Which tests a degenerate limit $ω \to 0$ ? ::: Ex 5 (line) Which produces a natural boundary condition? ::: Ex 6 ( $y^{'} (1) = 0$ ) Which uses a Lagrange multiplier? ::: Ex 7 (circle arc) Which is a real-world word problem? ::: Ex 8 (brachistochrone) Which springs the sign trap / higher derivative? ::: Ex 9 (Euler–Poisson, $y^{(4)} = 0$ )